Geodesics on Regular Polyhedra with Endpoints
at the Vertices

In a recent work of Davis et al.
([Davis et al.2016]), the authors consider geodesics on regular polyhedra
which begin and end at vertices (and do not touch other vertices).
The cases of regular tetrahedra and cubes are considered. The
authors prove that (in these cases) a geodesic as above never begins
at ends at the same vertex and compute the probabilities with which
a geodesic emanating from a given vertex ends at every other vertex.
The main observation of the present article is that there exists a
close relation between the problem considered in Davis et al.
([Davis et al.2016]) and the
problem of classification of closed geodesics on regular polyhedra
considered in articles ([Fuchs and Fuchs2007]
and [Fuchs2014]). This approach
yields different proofs of result of Davis et al.
([Davis et al.2016]) and permits to
obtain similar results for regular octahedra and icosahedra (in
particular, such a geodesic never ends where it begins).

A geodesic
on a surface of a polyhedron is, by definition, a locally shortest
curve which may transversally intersect edges, but does not contain
vertices besides, possibly, the endpoints. A geodesic is straight
within the faces and at every intersection with an edge, the
opposite angles formed by the geodesic and the edge (in the two
faces attached to this edge) are equal. If a geodesic has two
endpoints at vertices, we call it a simple geodesic segment.
(A simple geodesic segment is allowed to have self-intersections.)

All the results of [Davis et al.2016] mentioned below are given with
proofs, which, at least in the case of the cube, are not the same as
in [Davis et al.2016].

Theorem 2.1.

([Davis et al.2016], Corollary 3.8)
A simple geodesic segment starting at some vertex of a regular
tetrahedron never ends at the same vertex and ends at the three
other vertices with equal probabilities.11Here and below,
speaking of the probability with which a geodesic segment with a
given starting point ends at some vertex, I mean the asymptotic
probability for the set of geodesics of bounded length.

Proof.

Consider the planar development of the regular tetrahedron with
vertices $a,b,c,d$ (see Fig. 1). A line segment in this plane starting at the leftmost point marked $a$ and ending at one of the vertices within the angle shown in
Fig. 1 has the form $p\,{\mathbf{u}}+q\,{\mathbf{v}}$ where $p$ and $q$ are non-negative integers. Namely, it ends at
a vertex marked $a$, if $p$ and $q$ are both even, at a vertex
marked $b$, if $p$ is odd and $q$ is even, at a vertex marked $c$,
if $p$ is even and $q$ is odd, and at a vertex marked $d$, if $p$
and $q$ are both odd. It corresponds to a simple geodesic segment if
and only if $p$ and $q$ are relatively prime; in particular, if it
ends at $a$, then it does not correspond to any simple geodesic
segment. Theorem follows.

Theorem 3.1.

A simple geodesic segment starting at some vertex of a regular
octahedron never ends at the same vertex. It ends at the opposite
vertex with the probability $\displaystyle\frac{1}{4}$ and at the each
of the other vertices with the probability
$\displaystyle\frac{3}{16}$.

Proof.

Figure 2 shows a regular octrahedron (left) and its
planar development. The latter is multivalued in the sense that
every vertex $a,b,c$ is, actually, two vertices ($a$ represents $a$
and $a^{\prime}$, etc.), every edge represents four edges ($ab$ represents
$ab,a^{\prime}b,ab^{\prime}$, and $a^{\prime}b^{\prime}$, etc.), and every triangular face
represents eight triangular faces. Thus every line in the plane
emanating, say, from the leftmost vertex $a$ and not following the
edges represents a (polygonal) line on the surface of the
octahedron, but following this line we may need to replace $a$ by
$a^{\prime}$, $b$ by $b^{\prime}$, and/or $c$ by $c^{\prime}$.

Figure 2: A regular octahedron and its (multivalued) planar
development

For positive relatively prime $p,q$, a vector $p\,{\mathbf{u}}+q\,{\mathbf{v}}$ represents a simple geodesic segment on the surface
ending at $a$ or $a^{\prime}$, if $p-q\equiv 0\bmod 3$, ending at $b$ or $b^{\prime}$,
if $p-q\equiv 1\bmod 3$ and ending at $c$ or $c^{\prime}$, if
$p-q\equiv-1\bmod 3$. The most importrant thing we need to prove is
that it never ends at $a$.

Let $p>q>0,\,{\rm GCD}(p,q)=1,\,p\equiv q\bmod 3$. Figure
3 shows a parallelogram spanned by $p\,{\mathbf{u}}$ and $q\,{\mathbf{v}}$ (on the picture, $p=13,q=7$). We consider
the diagonal segment covered by triangles of the tiling. The
vertices are labeled according to the following rules. The left
lower triangle is $abc$; if two triangles share a side, then the two
vertices not on this side are labeled by the same letter, one with
prime, one without prime, like $a$ and $a^{\prime}$, $a^{\prime}$ and $a$, etc. We
are interested in vertices labeled by $a$ or $a^{\prime}$. The form a
sequence $a,a^{\prime},a,a^{\prime},\dots$ where the neighbors are vertices of the
triangles sharing a side not containing these vertices. These sides
are marked (encircled) on the picture. We need to show that the
number of marked sides is odd. It is 13 on the picture. Let us
consider the general case.

Figure 3: The segment $p\,{\mathbf{u}}+q\,{\mathbf{v}}$
covered by the triangles of the tiling

Let $r=\displaystyle\frac{q}{p-q}$. The following segments are marked
(coordinates are given in the basis $\{{\mathbf{u}},{\mathbf{v}}\}$):

[the last segment is, actually, $(p-1,q)-(p,q-1)$]. In this listing
of segments, we see the alternating of groups of parallel segments
and stair-like chains. The groups of parallel segments contain,
respectively, $[r]+1,[4r]+[2r]+1,\dots,[(p-q-2)r]-(p-q-4)r]+1,q-[(p-q-1)r]$ items; the stair-like chains contain, respectively,
$2([2r]-[r])+1,2([5r]-[4r])+1,\dots,2([(p+q-1)r]-[(p+q-2)r]+1$items. The total is

where the number of 1’s is $\displaystyle\frac{p-q}{3}$. A simple
computation shows that this sum is $\displaystyle{2\left(\left[\frac{p}{3}\right]+\left[\frac{q}{3}\right]\right)}+1$, which is certainly odd.

We arrive at the following result. No simple geodesic segment
emanating from $a$ ends at $a$. If this segment is determined by
relatively prime $p,q$, then it ends at $a^{\prime}$ if and only if $p\equiv q\bmod 3$; asymptotically, the probability of this event is
$\displaystyle\frac{1}{4}$. The remaining $\displaystyle\frac{3}{4}$ is
equally (because of the symmetry) distributed between $b,b^{\prime},c,$ and
$c^{\prime}$.

We begin this section with formulating some results from [Fuchs and Fuchs2007]
and [Fuchs2014].

We label the vertices of the cube as shown in Fig. 4,
left. An arbitrary curve on the surface of the cube not passing
through the vertices may be presented as a curve in the plane
furnished with the standard square lattice, and there arises a
labeling of the vertices along this curve. Important remark:
vertices of the lattice whose coordinates have the same parity get
the labels from the set $\{a,b^{\prime},c,d^{\prime}\}$, and vertices whose
coordinates have different parities get the labels from the set
$\{a^{\prime},b,c^{\prime},d\}$ (marked in Fig. 4).

Figure 4: The cube and a planar development of the segment
$p\cdot{\mathbf{u}}+q\cdot{\mathbf{v}}$ with
$p=3,q=2$

Straight segments in the planar development correspond to geodesics
on the surface of the cube. A segment $[(a,b),(a+p,b+q)]$ with
integer $a,b$ and relatively prime $p,q$ corresponds to a simple
geodesic segment on the cube joining two vertices of the cube, and
all such geodesic segments can be obtained in this way.

If we shift the segment to a parallel segment starting at the point
$(\varepsilon,0)$ with a small $\varepsilon>0$, then we get a
geodesic on the cube, which may be (and, actually, always is) not
closed: the segment in Fig. 4, right, shifted a
little bit to the right, corresponds to a geodesic starting near $a$
and ending near $a^{\prime}$. To get a closed geodesic, we need to repeat
the segment on the plane a certain (minimal) number $n_{C}(p,q)$ of
times.

Theorem 4.1.

([Fuchs and Fuchs2007], Theorem 4.4)

(1)

For all
$p,q,\ n_{C}(p,q)=2,3,$
or
$4;$

(2)

if
$p$
and
$q$
are both odd, then
$n_{C}(p,q)=3;$

(3)

if one of
$p,q$
is even, then
$n_{C}(p,q)=2$
or
4
.

To formulate a necessary result from [Fuchs2014], we need some
notation. Let $\mathcal{S}$ be the set of pairs $(p,q)$ of integers
such that $p$ is even, $q$ is odd, and $p,q$ do not have common
divisors $>1$ [thus, for example, $(0,1),(2,1)\in\mathcal{S}$, but
$(0,3)\notin\mathcal{S}$]. The group $\Gamma_{2}$ of integer $2\times 2$ matrices congruent to the identity modulo 2 acts transitively in
$\mathcal{S}$; it is known that $\Gamma_{2}$ is a free group with
generators $\displaystyle{A=\left[\begin{array}[]{cc}1&2\\
0&1\end{array}\right],B=\left[\begin{array}[]{cc}1&0\\
2&1\end{array}\right]}$. Let $H\subset\Gamma_{2}$ be the subgroup
generated by $A^{2},B^{2},ABA$ and $BAB$. It is proved in [Fuchs2014]
that $H$ has index 3 in $\Gamma_{2}$ and not normal.

Theorem 4.2.

[[Fuchs2014], Theorem 2.2, Part
(3)] The group $H$ has two orbits in $\mathcal{S}$, and these orbits
are $\{n_{C}(p,q)=2\}$ and $\{n_{C}(p,q)=4\}$. The asymptotic size of
the second of these orbits is twice the asymptotic size of the first
one.

We will need here some additional facts. For every pair $(p,q)$ of
relatively prime integers the segment $p\,{\mathbf{u}}+q\,{\mathbf{v}}$ determines a simple geodesic segment on the surface of the cube
starting at $a$. Denote by $V(p,q)$ the endpoint of this geodesic.
For example, Fig. 4 shows that $V(3,2)=a^{\prime}$. A small
table of values of the function $V$ is shown in Fig.
5.

Figure 5: The function $V(p,q)$

The function $V$ has a lot of symmetries, both visible and hidden.
First of all, if, for a given $(p,q)$, the pair $(p^{\prime},q^{\prime})$ is one of
$(-p,q),\,(p,-q),\,(q,p)$ or $(-q,-p)$, then $V(p^{\prime},q^{\prime})$ is easily
related to $V(p,q)$. For example, if $q>0$, then $V(-p,q)$ is
obtained from $V(p,q)$ by the transformation $a^{\prime}\leftrightarrow b,\,c\leftrightarrow d^{\prime}$ $\Big($indeed, all we need for this transition,
is the replacement of the square
$\overline{\underline{\left|\begin{array}[]{cc}d&c\\
a&b\end{array}\right|}}$ by the square
$\overline{\underline{\left|\begin{array}[]{cc}d&d^{\prime}\\
a&a^{\prime}\end{array}\right|}}\Big)$ etc. It is important that all these
transformations preserve $a$ (and $c^{\prime}$).

To describe some other symmetries, consider the following two
matrices from $H$: $C=B^{2}=\left[\begin{array}[]{cc}1&0\\
4&1\end{array}\right],D=-AB^{-1}A=\left[\begin{array}[]{cc}3&4\\
2&3\end{array}\right]$.

Proof.

Essentially, this theorem is proved in [Fuchs2014]. The proof is
contained in Fig. 6 [borrowed from [Fuchs2014]].

Figure 6: To proof of Lemma
4.3

We can replace the standard fundamental square of the lattice by one
of the parallelograms $(0,0),(1,4),(1,5),(0,1)$ or
$(0,0),(3,2),(7,5),(4,3)$. The labeling of vertices shown
in Fig. 6 provides labels for the vertices of
these two parallelogram. In the left diagram, it is again $abcd$,
and this proves Part (1). In the right diagram, it is $aa^{\prime}b^{\prime}b$ which
means that the whole labeling is transformed by the rotation of the
cube which maps the face $abcd$ into the face $aa^{\prime}b^{\prime}b$. This is the
rotation of the cube by $120^{\circ}$ around the diagonal $c^{\prime}a$, as
described in Part (2).

Now we formulate the main result of this section.

Theorem 4.4.

Consider a simple geodesic segment on the cube emanating from the
vertex $a$; suppose that it corresponds to some relatively prime
$p,q$.

(1)

([Davis et al.2016], Corollary 5.15, Theorem 5.17) The vertex
$V(p,q)$ cannot be the same as $a$; it is one of $a^{\prime},b,d$ with the
probability $\displaystyle\frac{4}{27}$ for each; one of $b^{\prime},c,d^{\prime}$
with the probability $\displaystyle\frac{1}{9}$ for each and it is $c^{\prime}$
with the probability $\displaystyle\frac{2}{9}$.

(2)

The vertex
$V(p,q)$
is one of
$a^{\prime},b,d$
, if
$n_{C}(p,q)=4$
, is
one of
$b^{\prime},c,d^{\prime}$
, if
$n_{C}(p,q)=3$
, and is
$c^{\prime}$
, if
$n_{C}(p,q)=2$
.

Proof.

As in the proof of Theorem 3.1, the main thing we need to
prove is that a simple geodesic segment beginning at a vertex of the
cube never ends at the same vertex. It is clear from the remark in
the beginning of the section that if it the endpoint of the simple
geodesic segment coincides with the beginning, then this segment
corresponds to the pair $(p,q)$ with $p,q$ being both odd.

Return to Theorem 4.4. Notice that for arbitrary
relatively prime odd $p,q$ there exists a sequence of
transformations described before and in Lemma 4.3 which
reduce the pair $(p,q)$ to $(1,1)$. Indeed, combining the
transformation $C$ with sign changes and swapping of coordinates, we
can reduce the general case to the case of positive $p,q$ with
$p<q<2p$ [besides the case $(p,q)=(1,1)$]; and this transformation
does not increase the minimum of absolute values $|p|,|q|$. Then we
apply $D$; $(p,q)$ becomes $(4p-3q,3p-2q)$, and $2p<2q<4p\Rightarrow-p<3p-2q<q$, or $|3p-2q|<p$; thus the minimum of $|p|,|q|$
decreases. Repeating this procedure sufficiently many times, we
arrive at $(p,q)=(1,1)$. Thus an arbitrary $(p,q)$ can be obtained
from $(1,1)$ by a chain of inverse transformation, and, since
$V(1,1)=c$ and no one of our transformations connects $a$ with
anything else, we can conclude that $V(p,q)\neq a$.

Finally, consider an arbitrary simple geodesic segment $\sigma$
starting at $a$; let its planar development is $p\,{\mathbf{u}}+q\,{\mathbf{v}}$. Take the geodesic segment $\sigma^{\prime}$ parallel to
$\sigma$ and starting at a point of the edge $ab$ close to $a$. If
$\sigma$ ends at $b,d,$ or $a^{\prime}$ then repeat $\sigma^{\prime}$ 4 times; if
$\sigma$ ends at $c,b^{\prime}$, or $d^{\prime}$, then repeat $\sigma^{\prime}$ 3 times; and
if $\sigma$ ends at $c^{\prime}$, then repeat $\sigma^{\prime}$ 2 times. Obviously,
we get a geodesic on the cube ending at a point of one of the edges
$ab,ad,$ or $aa^{\prime}$. But in the last two cases, to get a closed
geodesic, we need to repeat already repeated 4, or 3, or 2 times
$\sigma^{\prime}$ 3 more times, which would mean that $n_{C}(p,q)=12$ or 9, or
6, in contradiction to Part (3) of Theorem 4.1. Thus, the
geodesic $\sigma^{\prime}$ repeated 4, or 3, or 2 times is closed and we
conclude that if $V(p,q)$ is $b,d$, or $a^{\prime}$, then $n_{C}(p,q)=4$, if
$V(p,q)$ is $c,b^{\prime}$ or $d^{\prime}$, then $n_{C}(p,q)=3$ and if $V(p,q)$ is
$c^{\prime}$, then $n_{C}(p,q)=2$. Theorem 4.4 follows.

We label the 12 vertices of the regular icosahedron as shown in
Fig. 7. Notice that, here prime means “opposite”:
the vertex $a^{\prime}$ is opposite to $a$, the vertex $b^{\prime}$ is opposite to
$b$, etc.

Figure 7: Vertices of the icosahedron

As in the cases of the tetrahedron and the octahedron, geodesics on
the surface of the icosahedron (not passing through the vertices)
are presented by straight lines in the plane furnished by the
standard triangular tiling. A non-self-repeating closed geodesic
corresponds to the segment $p\,{\mathbf{u}}+q\,{\mathbf{v}}$ with
relatively prime integer $p,q$ repeated a certain number $n_{I}(p,q)$
of times.

Theorem 5.1.

([Fuchs and Fuchs2007], Theorem 6.1)
The number $n_{I}(p,q)$ takes values $2,3$, and 5.

The following informations of the function $n_{I}$ are contained in
[Fuchs2014]. Let ${\mathcal{S}}=\{(p,q)\in{\mathbb{Z}}\times{\mathbb{Z}}\mid{\rm GCD}(p,q)=1\}/(%
p,q)\sim(-p,-q)$. The group $PSL(2,{\mathbb{Z}})=SL(2,{\mathbb{Z}})/\{\pm I\}$ transitively acts in $\mathcal{S}$.
Let $H$ be the subgroup of $PSL(2,{\mathbb{Z}})$ generated by
$K=\left[\begin{array}[]{cc}0&-1\\
1&\phantom{-}1\end{array}\right],L=\left[\begin{array}[]{cc}-4&1\\
-1&0\end{array}\right],$ and
$M=\left[\begin{array}[]{cc}4&-3\\
3&-2\end{array}\right]$.

Theorem 5.2.

([Fuchs2014], Theorem 3.1)

(1)

The group
$H$
has index
10
in
$PSL(2,{\mathbb{Z}}).$

(2)

The group
$H$
has three orbits in
$\mathcal{S}$
, and these
orbits are
$\{n_{I}(p,q)=2\},\{n_{I}(p,q)=3\},$
and
$\{n_{I}(p,q)=5\}$
.

(3)

The asymptotic sizes of these three orbits are related as
$2{:}3{:}5$
.

As in the case of the cube, we need some enhancement of Theorem
5.2. For relatively prime $p,q$, let $V(p,q)$ be the
vertex of the icosahedron, which is the endpoint of the simple
geodesic segment beginning at $a$ and corresponding to the planar
segment $p\,{\mathbf{u}}+q\,{\mathbf{v}}$.

Proof.

As in the proof of Lemma 4.3, the transformations
consist in replacements of the fundamental triangle of the tiling as
shown in Fig. 8 [mostly borrowed from [Fuchs2014]].
We see that the transformations in Parts (1)–(3) of lemma
correspond to the replacement of the fundamental triangle $abc$
(shadowed in Fig. 8) by, respectively, triangles
$abc,acd,$ and $afb$ (drawn in boldface lines in Fig.
8). Lemma follows.

The following statement is the main result of this section.

Theorem 5.4.

Consider a simple geodesic segment on the icosahedron emanating from
the vertex $a$ and ending at some vertex $h$; suppose that it
corresponds to some relatively prime $p,q$.

(1)

The vertex
$h$
cannot be
$a$
.

(2)

The vertex
$h$
is one of the vertices
$b,c,d,e,f$
with the
probabilitiy
$\displaystyle\frac{1}{10}$
for each; it is one of the
vertices
$b^{\prime},c^{\prime},d^{\prime},e^{\prime},f^{\prime}$
with the same probability
$\displaystyle\frac{3}{50}$
for each; and it is
$a^{\prime}$
with the
probability
$\displaystyle\frac{1}{5}$
.

(3)

The vertex
$h$
is one of
$b,c,d,e,f$
, if
$n_{I}(p,q)=5$
, it is
one of
$b^{\prime},c^{\prime},d^{\prime},e^{\prime},$
$f^{\prime}$
, if
$n_{I}(p,q)=3$
, and it is
$a^{\prime}$
, if
$n_{!}(p,q)=2$
.

Proof.

Add to Lemma 5.3 that $V(-p,-q)$ is also obtained from
$V(p,q)$ by a rotation, this time $b\mapsto e\mapsto c\mapsto f\mapsto d\mapsto b,\,b^{\prime}\mapsto e^{\prime}%
\mapsto c^{\prime}\mapsto f^{\prime}\mapsto d^{\prime}\mapsto b^{\prime}$. Consider the group $\widetilde{H}\subset SL(2;{\mathbb{Z}})$ generated by the matrices $-I,K,L,M$. The action of the group
$\widetilde{H}$ in $\widetilde{\mathcal{S}}$ has three orbits, the
inverse images the orbits of $H$ in $\mathcal{S}$ with resopect to
the projection $\widetilde{\mathcal{S}}\to\mathcal{S}$. Lemma
5.3, supplemented by the last remark, shows that the
sets $T_{A}=\{(p,q)\mid V(p,q)\in A$ with $A$ being one of the four
sets $\{a\},\,\{b,c,d,e,f\},\,\{b^{\prime},c^{\prime},d^{\prime},e^{\prime},f^{%
\prime}\},\,\{a^{\prime}\}$ are
invariant with respect to $\widetilde{H}$. Since the three orbits are
represented by the pairs $(1,1),(1,2),(1,4)$ and $V(1,1)=c^{\prime},\,V(1,2)=a^{\prime},\,V1,4)=d$, We conclude that $T_{\{a\}}$ is empty, while
the three other sets $T_{A}$ are the three orbits of $\widetilde{H}$,
and Theorem 5.2 yields identifications $T_{\{b,c,d,e,f\}}=n_{I}^{-1}(5),\,T_{\{b^{\prime},c^{\prime},d^{\prime},e^{%
\prime},f^{\prime}\}}=n_{I}^{-1}(3),\,T_{\{a^{\prime}\}}=n_{!}^{-1}(2).$ Theorem follows.

I cannot say much about this case. Still it seems very likely that
on the surface of the dodecahedron there exist simple geodesic
segments which begin and end at the same vertex. By means of
computer experiments, I have found a variety of likely examples of
this. The shortest one is shown in Fig. 9. It is a
22-gonal geodesic emanating from the vertex $A$ under the angle
$\alpha=\tan^{-1}\displaystyle\frac{31\sqrt{3-\phi}}{57\phi-90}\approx 86.5^{\circ}$
with the side ($\phi=1.618...$ is the golden ratio). According to
the computer, the distance between the 22nd edge and $A$ does not
exceed $10^{-14}$ (we assume that the length of the side is 1). In
the same time, the distances between the other edges and other
vertices are never less than 0.02. This means that in the unlikely
event that the 22nd edge does not hit $A$, we can make it ending at
$A$ by a small perturbation of $\alpha$.